If i am moving a charged particle in a circular motion and then applying the magnetic field perpendicular to this circular motion ,will the charge experience a force. Note that magnetic line produced by charge and the applied magnetic field are in the same direction.


1 Answer 1


Yes; instantaneously, the particle will always have a velocity perpendicular to the magnetic field, so there will be a Lorentz force \$q\vec{v}\times \vec{B}\$, regardless of what extra external forces you apply.

Any extra forces you apply externally to maintain a circular orbit perpendicular to the magnetic field will have the effect of either maintaining a larger or smaller orbit than the gyroradius.

If you think about it, there are only 2 non-zero options for this force which will maintain a perfectly circular orbit perpendicular to a uniform magnetic field:

  1. It is perpendicular to the particle pointed towards the center of the orbit. This case is trying to maintain a smaller orbit than the natural gyroradius.
  2. It is perpendicular and pointed away from the center of the orbit. This case is trying to maintain a larger orbit than the natural gyroradius.

Another simpler way to resolve the question is to notice that \$\frac{d\vec{v}}{dt}\$ is not zero for any circular orbit (it rotates); this is the definition of acceleration, and the only way to produce an acceleration is by applying a force.

  • \$\begingroup\$ If we consider that particle is moving in the circular motion due to the magnetic field then this circular motion will again experience another force from the same magnetic field which created it ,but no text books mentioned this phenomenon. Does it happen this way? \$\endgroup\$
    – FranCliP
    Jun 5, 2015 at 18:29
  • \$\begingroup\$ Particles can't move in a circular motion without an external force; Newton's 1st laws of motion prohibits this. The force which causes circular motion in this case is the magnetic field force. \$\endgroup\$ Jun 5, 2015 at 21:28
  • \$\begingroup\$ You are right , but the circular motion caused by the magnetic field is perpendicular to the magnetic field so again there should be a lorentz force acting on the particle moving in the circular motion , so what will be the final equilibrium motion of the particle , this is my doubt? \$\endgroup\$
    – FranCliP
    Jun 6, 2015 at 5:09
  • \$\begingroup\$ The true particle motion is perfect circle. You are trying to apply a Lorentz force on the entire circular orbit at once, but you can't do that. You must look at each instant in time to calculate the force due to the magnetic field. There is exactly 1 Lorentz force at any instant in time, and it acts to keep the particle in an orbit. If you integrate the Lorentz force over the entire orbit (i.e. get a work), you'll notice that the work done is exactly 0. \$\endgroup\$ Jun 6, 2015 at 6:26
  • \$\begingroup\$ If we use the spherical coordinates , if magnetic field is in z direction and the velocity of the particle moving in circular motion is in the theta direction , then there should be a force in the r vector direction pointing away or towards the center of the circle . Am i right ? \$\endgroup\$
    – FranCliP
    Jun 6, 2015 at 8:33

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